How To Find Locus Of A Point
Eugene is a qualified control/instrumentation engineer Bsc (Eng) and has worked as a programmer of electronics & software for SCADA systems.
The Parabola, a Mathematical Function
In this tutorial you'll learn about a mathematical role called the parabola. We'll cover the definition of the parabola get-go and how information technology relates to the solid shape called the cone. Next we'll explore dissimilar means in which the equation of a parabola can exist expressed. Likewise covered will be how to work out the maxima and minima of a parabola and how to observe the intersection with the x and y axes. Finally we'll discover what a quadratic equation is and how you can solve it.
Definition of a Parabola
"A locus is a curve or other figure formed past all the points satisfying a detail equation."
One way we tin ascertain a parabola is that it is the locus of points that are equidistant from both a line called the directrix and a point chosen the focus. So each point P on the parabola is the same distance from the focus as information technology is from the directrix every bit y'all can see in the animation beneath.
We notice also that when 10 is 0, the distance from P to the focus equals the altitude from the vertex to the directrix. And so the focus and directrix are equidistant from the vertex.
A Parabola is a Conic Section
Another mode of defining a parabola
When a plane intersects a cone, we become different shapes or conic sections where the aeroplane intersects the outer surface of the cone. If the plane is parallel to the bottom of the cone, nosotros just get a circumvolve. As the angle A in the animation below changes, information technology eventually becomes equal to B and the conic section is a parabola.
Equations of Parabolas
At that place are several ways we can limited the equation of a parabola:
- As a quadratic function
- Vertex form
- Focus course
We'll explore these later, but beginning let'southward await at the simplest parabola.
The Simplest Parabola y = x²
The simplest parabola with the vertex at the origin, point (0,0) on the graph, has the equation y = x².
The value of y is merely the value of x multiplied past itself.
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Graph of y = x²
Let's Give 10 a Coefficient!
The simplest parabola is y = 10two but if we give x a coefficient, we tin generate an infinite number of parabolas with dissimilar "widths" depending on the value of the coefficient ɑ.
So lets brand y = ɑx2
In the graph beneath, ɑ has various values. Notice that when ɑ is negative, the parabola is "upside down". We'll discover more almost this later. Remember the y = ɑx2 form of the equation of a parabola is when its vertex is at the origin.
Making ɑ smaller results in a "wider" parabola. If we make ɑ bigger, the parabola gets narrower.
Turning the Simplest Parabola on Its Side
If we plough the parabola y = x2 on its side, we get a new function y2 = x or x = y2. This merely ways nosotros tin retrieve of y as being the independent variable and squaring it gives us the corresponding value for x.
So:
When y = two, 10 = y2 = 4
when y = iii, x = ytwo = 9
when y = 4, x = y2 = xvi
and and then on...
Just like the example of the vertical parabola, nosotros can once more add a coefficient to y2.
So we take x = ɑy2
Vertex Form of a Parabola Parallel to Y Centrality
One fashion we can express the equation of a parabola is in terms of the coordinates of the vertex. The equation depends on whether the axis of the parabola is parallel to the x or y axis, but in both cases, the vertex is located at the coordinates (h,1000). In the equations, ɑ is a coefficient and can have whatsoever value.
When the centrality is parallel to y axis:
y = ɑ(x - h)2 + k
if ɑ = 1 and (h,k) is the origin (0,0) we go the simple parabola we saw at the commencement of the tutorial:
y = ane(x - 0)2 + 0 = x2
When the centrality is parallel to the x axis:
x = ɑ(y - h)2 + k
Find that this doesn't requite u.s.a. any information about the location of the focus or directrix.
Equation of a Parabola in Terms of the Coordinates of the Focus
Another way of expressing the equation of a parabola is in terms of the coordinates of the vertex (h,thousand) and the focus.
Nosotros saw that:
y = ɑ(x - h)2 + k
Using Pythagoras's Theorem we can testify that the coefficient ɑ = 1/4p, where p is the distance from the focus to the vertex.
When the axis of symmetry is parallel to y axis:
Substituting for ɑ = one/4p gives us:
y = ɑ(x - h)2 + k = 1/(4p)(x - h)two + k
Multiply both sides of the equation past 4p:
4py = (ten - h)two + 4pk
Rearrange:
4p(y - thousand) = (ten - h)ii
or
(x - h)ii = 4p(y - yard)
Similarly:
When the axis of symmetry is parallel to x axis:
A similar derivation gives us:
(y - k)2 = 4p(x - h)
Example:
Find the focus for the simplest parabola y = xii
Reply:
Since the parabola is parallel to the y axis, we use the equation we learned well-nigh above
(x - h)2 = 4p(y - 1000)
First detect the vertex, the point where the parabola intersects the y axis (for this uncomplicated parabola, we know the vertex occurs at x = 0)
So ready x = 0, giving y = ten2 = 0two = 0
and therefore the vertex occurs at (0,0)
But the vertex is (h,k), therefore h = 0 and k = 0
Substituting for the values of h and one thousand, the equation (x - h)ii = 4p(y - 1000) simplifies to
(x - 0)2 = 4p(y - 0)
giving united states of america
102 = 4py
Now compare this to our original equation for the parabola y = x2
We can rewrite this every bit xtwo = y, but the coefficient of y is 1, so 4p must equal 1 and p = i/iv.
From the graph above, we know the coordinates of the focus are (h, k + p), so substituting the values nosotros worked out for h, k and p gives us the coordinates of the vertex as
(0, 0 + ane/4) or (0, 1/4)
A Quadratic Function is a Parabola
Consider the function y = ɑx2 + bx + c
This is called a quadratic role because of the square on the x variable.
This is some other style we can express the equation of a parabola.
How to Determine Which Direction a Parabola Opens
Irrespective of which form of equation that is used to depict a parabola, the coefficient of xtwo determines whether a parabola will "open up" or "open down". Open up means that the parabola will accept a minimum and the value of y will increase on both sides of the minimum. Open downward ways it volition have a maximum and the value of y decreases on both sides of the max.
- If ɑ is positive, the parabola volition open upwardly
- If ɑ is negative the parabola will open down
Parabola Opens Upward or Opens Downwardly
How to Discover the Vertex of a Parabola
From uncomplicated calculus nosotros tin deduce that the max or min value of a parabola occurs at 10 = -b/2ɑ
Substitute for 10 into the equation y = ɑx2 + bx + c to get the corresponding y value
And so y = ɑx2 + bx + c
= ɑ(-b/2ɑ)2 + b(-b/2ɑ) + c
= ɑ(b2/4ɑtwo) - b2/2ɑ + c
Collecting up the btwo terms and rearranging
= b2 (ane/4ɑ - i/2ɑ) + c
= - b2/4ɑ + c
= c -b2/4a
So finally the min occurs at (-b/2ɑ, c -b2/4ɑ)
Example:
Find the vertex of the equation y = 5x2 - 10x + 7
- The coefficient a is positive, so the parabola opens up and the vertex is a minimum
- ɑ = 5, b = -10 and c = 7, then the x value of the minimum occurs at 10 = -b/2ɑ = - (-10)/(two(5)) = 1
- The y value of the min occurs at c - btwo/4a. Substituting for a, b and c gives us y = vii - (-x)2 / (4(five)) = 7 - 100/xx = 7 - 5 = 2
So the vertex occurs at (i,2)
How to Discover the X-Intercepts of a Parabola
A quadratic function y = ɑxtwo + bx + c is the equation of a parabola.
If nosotros set the quadratic part to zero, we become a quadratic equation
i.due east. ɑxtwo + bx + c = 0 .
Graphically, equating the function to cipher means setting a status of the function such that the y value is 0, in other words, where the parabola intercepts the ten axis.
The solutions of the quadratic equation allow us to notice these ii points. If there are no real number solutions, i.e. the solutions are imaginary numbers, the parabola doesn't intersect the 10 axis.
The solutions or roots of a quadratic equation are given by the equation:
ten = -b ± √(b2 -4ac) / 2ɑ
Finding the Roots of a Quadratic Equation
Example one: Find the x-axis intercepts of the parabola y = 3xtwo + 7x + ii
Solution
- y = ɑx2 + bx + c
- In our example y = 3x2 + 7x + ii
- Identify the coefficients and constant c
- And then ɑ = three, b = 7 and c = 2
- The roots of the quadratic equation 3xtwo + 7x + 2 = 0 are at x = (-b ± √(b2 - 4ɑc)) / 2ɑ
- Substitute for ɑ, b and c
- The starting time root is at x = (-7 + √(vii2 - four ten 3 x 2)) / (2 x 3) = -1/3
- The second root is at (-7 - √(7two - 4 x three x ii)) / (2 x 3) = -ii
- So the x centrality intercepts occur at (-ii, 0) and (-1/3, 0)
Example 2: Find the x-axis intercepts of the parabola with vertex located at (4, 6) and focus at (4, 3)
Solution
- The equation of the parabola in focus vertex form is (x - h)ii = 4p(y - k)
- The vertex is at (h,k) giving united states h = iv, 1000 = 6
- The focus is located at (h, k + p). In this example the focus is at (4, 3) so k + p = iii. But k = vi then p = iii - half-dozen = -three
- Plug the values into the equation (10 - h)two = 4p(y - k) then (x - 4)two = 4(-3)(y - 6)
- Simplify giving (x - iv)2 = -12(y - 6)
- Expand out the equation gives us 10two - 8x + 16 = -12y + 72
- Rearrange 12y = -xtwo + 8x + 56
- Giving y = -i/12x2 + 2/3x + xiv/3
- The coefficients are a = -1/12, b = 2/3, c = xiv/3
- The roots are at (-2/3 ± √((two/3)ii - four(-1/12)(14/3)))/(2(-1/12)
- This gives usa x = -4.49 approx and x = 12.49 approx
- So the x axis intercepts occur at (-4.49, 0) and (12.49, 0)
How to Find the Y-Intercepts of a Parabola
To detect the y-axis intercept (y-intercept) of a parabola, we set up x to 0 and summate the value of y.
Example 3: Detect the y-intercept of the parabola y = 6x2 + 4x + 7
Solution:
y = 6x2 + 4x + 7
Set x to 0 giving
y = 6(0)2 + iv(0) + 7 = vii
The intercept occurs at (0, vii)
Summary of Parabola Equations
Equation Blazon | Axis Parallel to Y-Axis | Centrality Parallel to 10-Axis |
---|---|---|
Quadratic Office | y = ɑx² + bx + c | x = ɑy² + by + c |
Vertex Class | y = ɑ(x - h)² + k | x = ɑ(y - h)² + k |
Focus Form | (ten - h)² = 4p(y - one thousand) | (y - k)² = 4p(x - h) |
Parabola with Vertex at the Origin | ten² = 4py | y² = 4px |
Roots of a parabola parallel to y axis | x = -b ± √( b² -4ɑc)/2ɑ | |
Vertex occurs at | (-b/2ɑ, c -b2/4ɑ) |
How the Parabola is Used in the Real World
The parabola isn't just confined to math. The parabola shape appears in nature and we apply it in science and engineering science because of its properties.
- When you kick a ball into the air or a projectile is fired, the trajectory is a parabola
- The reflectors of vehicle headlights or flashlights are parabolic shaped
- The mirror in a reflecting telescope is parabolic
- Satellite dishes are in the shape of a parabola equally are radar dishes
- Microphones for listening to fain sounds tin can have parabolic reflectors fitted to collect audio. The microphone is located at the focus to collect sound.
For radar dishes, satellite dishes and radio and optical telescopes, one of the parabola's properties is that a ray of electromagnetic radiations parallel to its axis will be reflected towards the focus. Conversely in the case of a headlight or torch, light coming from the focus will be reflected off the reflector and travel outwards in a parallel beam.
Related reading: Deriving Projectile Movement Equations
Parabolic Reflector
Acknowledgements
All graphics were created using GeoGebra Classic.
© 2019 Eugene Brennan
Srinjan Sanyal on May 27, 2020:
the content is very good.
rafia from lahore pakistna on September 29, 2019:
hey y'all are a dainty teacher! i previously did not know the open up up or down graph ..but now i do ..thank yous!
Source: https://owlcation.com/stem/How-to-Understand-the-Equation-of-a-Parabola-Directrix-and-Focus
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